Question

1379. Find a Corresponding Node of a Binary Tree in a Clone of That Tree

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Description

Given two binary trees original and cloned and given a reference to a node target in the original tree.

The cloned tree is a copy of the original tree.

Return _a reference to the same node_ in the cloned tree.

Note that you are not allowed to change any of the two trees or the target node and the answer must be a reference to a node in the cloned tree.

 

Example 1:

Input: tree = [7,4,3,null,null,6,19], target = 3
Output: 3
Explanation: In all examples the original and cloned trees are shown. The target node is a green node from the original tree. The answer is the yellow node from the cloned tree.

Example 2:

Input: tree = [7], target =  7
Output: 7

Example 3:

Input: tree = [8,null,6,null,5,null,4,null,3,null,2,null,1], target = 4
Output: 4

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• The values of the nodes of the tree are unique.
• target node is a node from the original tree and is not null.

 

Follow up: Could you solve the problem if repeated values on the tree are allowed?

Solutions

Solution 1: DFS

We design a function $dfs(root1, root2)$, which performs DFS traversal simultaneously in trees $root1$ and $root2$. When traversing to a node, if this node happens to be $target$, then we return the corresponding node in $root2$. Otherwise, we recursively search for $target$ in the left and right subtrees of $root1$ and $root2$, and return the result that is not empty.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes in the tree.

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def getTargetCopy(
    self, original: TreeNode, cloned: TreeNode, target: TreeNode
  ) -> TreeNode:
    def dfs(root1: TreeNode, root2: TreeNode) -> TreeNode:
      if root1 is None:
        return None
      if root1 == target:
        return root2
      return dfs(root1.left, root2.left) or dfs(root1.right, root2.right)

    return dfs(original, cloned)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */

class Solution {
  private TreeNode target;

  public final TreeNode getTargetCopy(
    final TreeNode original, final TreeNode cloned, final TreeNode target) {
    this.target = target;
    return dfs(original, cloned);
  }

  private TreeNode dfs(TreeNode root1, TreeNode root2) {
    if (root1 == null) {
      return null;
    }
    if (root1 == target) {
      return root2;
    }
    TreeNode res = dfs(root1.left, root2.left);
    return res == null ? dfs(root1.right, root2.right) : res;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
  TreeNode* getTargetCopy(TreeNode* original, TreeNode* cloned, TreeNode* target) {
    function<TreeNode*(TreeNode*, TreeNode*)> dfs = [&](TreeNode* root1, TreeNode* root2) -> TreeNode* {
      if (root1 == nullptr) {
        return nullptr;
      }
      if (root1 == target) {
        return root2;
      }
      TreeNode* left = dfs(root1->left, root2->left);
      return left == nullptr ? dfs(root1->right, root2->right) : left;
    };
    return dfs(original, cloned);
  }
};

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function getTargetCopy(
  original: TreeNode | null,
  cloned: TreeNode | null,
  target: TreeNode | null,
): TreeNode | null {
  const dfs = (root1: TreeNode | null, root2: TreeNode | null): TreeNode | null => {
    if (!root1) {
      return null;
    }
    if (root1 === target) {
      return root2;
    }
    return dfs(root1.left, root2.left) || dfs(root1.right, root2.right);
  };
  return dfs(original, cloned);
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left;
 *   public TreeNode right;
 *   public TreeNode(int x) { val = x; }
 * }
 */

public class Solution {
  private TreeNode target;

  public TreeNode GetTargetCopy(TreeNode original, TreeNode cloned, TreeNode target) {
    this.target = target;
    return dfs(original, cloned);
  }

  private TreeNode dfs(TreeNode original, TreeNode cloned) {
    if (original == null) {
      return null;
    }
    if (original == target) {
      return cloned;
    }
    TreeNode left = dfs(original.left, cloned.left);
    return left == null ? dfs(original.right, cloned.right) : left;
  }
}